to the integral \(I\) , we reduce it to a sum of the incomplete zeta functions. Furthermore, we expand these incomplete zeta functions making use of the formulas [1]}:
\(\begin{split}\int _a^\infty d\omega \frac{\omega ^{\nu -1}}{e^\omega -1}&=\Gamma (\nu )\zeta (\nu )-\sum _{n=-1}^\infty \frac{(-1)^n\zeta (-n)a^{\nu +n}}{\Gamma (n+1)(\nu +n)},\\\int _a^\infty d\omega \frac{\omega ^{\nu -1}}{e^\omega +1}&=(1-2^{1-\nu })\Gamma (\nu )\zeta (\nu )-\sum _{n=0}^\infty (1-2^{1+n})\frac{(-1)^n\zeta (-n)a^{\nu +n}}{\Gamma (n+1)(\nu +n)}.\end{split}\) 
